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Question 55110This question is from textbook
: determine vertex of y=x^2-8x+22
this is what I did (after lots of hair pulling)
y=x^2-8x+22
y-22=x^2-8x
y-22+16=x^2-8x+16
y-22+4=(x-4x)^2
y-18=(x-4x)^2
y=(x-4x)^2+18
answer: (-4,18)
Am I on the right track?
This question is from textbook
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! determine vertex of y=x^2-8x+22
You are almost on the right track.
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y=x^2-8x+22
x^2-8x+? = y-22+?
x^2-8x+16 = y-22+16
(x-4)^2 = y-6
This follows the form (x-h)^2=4p(y-k)
where (h,k) is the vertex.
Vertex is (4,6)
Cheers,
Stan H.
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